Eq. (REF ) is found by adapting a result known to Euler, see [1]}, for the recasting of a series
\(e_{n} = 1+\sum _{k=1}^{n} \prod _{j=1}^{k}\big (-E_{j}\big )z^k ,\) 
into
\(e_{n} = \cfrac{1}{1+\hbox{\huge $\mathrm {K}$}}^{n}_{j=1}\cfrac{\displaystyle E_{j}z}{\displaystyle 1-E_{j}z}\) 
.

It is straight forward to show the partial numerators and denominators \( A_n \)  and \( B_n \)  of \( e_{n} = \frac{A_n}{B_n} \) 
satisfy the recurrence relations
\(A_n & = b_nA_{n-1}+a_nA_{n-2} ,\\B_n & = b_nB_{n-1}+a_nB_{n-2} ,\) 

with \( b_0=0, a_1=1, b_1=1 \)  and for \( m \ge 1 \) , \( a_{m+1} = E_{m} z \)  and \( b_{m+1} = 1-E_{m}z \) .
The solutions to this recurrence give \( B_{n}=1 \)  and \( A_{n} \)  the right-hand side of (REF ).
